Question 1

In an experiment,a student brings up the rotational speed of a piece of laboratory apparatus to 30.0 rpm.She then allows the apparatus to slow down uniformly on its own,and counts 240 revolutions before the apparatus comes to a stop.The moment of inertia of the apparatus is known to be 0.0850 kg•m².What is the magnitude of the retarding torque on the apparatus? A) 0.0425 N • m B) 0.159 N • m C) 0.0787 N • m D) 0.000278 N • m E) 0.0000136 N • m

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Question 2

A uniform solid 5.25-kg cylinder is released from rest and rolls without slipping down an inclined plane inclined at 18° to the horizontal.How fast is it moving after it has rolled 2.2 m down the plane?
A) 4.3 m/s
B) 5.2 m/s
C) 3.0 m/s
D) 3.7 m/s
E) 2.6 m/s

Accepted Answer

We can use conservation of energy to solve this problem. The initial potential energy of the cylinder is converted to kinetic energy as it rolls down the inclined plane. The equation for conservation of energy is:

PEi = KEf
mgh = 1/2mv^2 + 1/2Iω^2

where m is the mass of the cylinder, h is the height of the inclined plane, v is the velocity of the cylinder, I is the moment of inertia of the cylinder, and ω is the angular velocity of the cylinder.

Since the cylinder is rolling without slipping, we can use the relationship between linear and angular motion: v = rω, where r is the radius of the cylinder.

We can express the moment of inertia in terms of the mass and radius: I = 1/2mr^2, since the cylinder is uniform.

Substituting these relationships into the conservation of energy equation and solving for v, we get:

v = (2gh/7)^0.5

Plugging in the values given in the problem, we get:

v = (2*9.81*sin18*2.2/7)^0.5 = 3.0 m/s

Therefore, the answer is C.

Question 3

A solid,uniform sphere of mass 2.0 kg and radius 1.7 m rolls from rest without slipping down an inclined plane of height 7.0 m.What is the angular velocity of the sphere at the bottom of the inclined plane?
A) 5.8 rad/s
B) 9.9 rad/s
C) 11 rad/s
D) 7.0 rad/s

Accepted Answer

The angular velocity can be found using energy conservation principles. The potential energy at the top is converted into translational and rotational kinetic energy at the bottom. The potential energy at the top is $mgh$, where $m = 2.0 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, and $h = 7.0 \, \text{m}$. The kinetic energy at the bottom is a sum of translational ($\frac{1}{2}mv^2$) and rotational ($\frac{1}{2}I\omega^2$) kinetic energies. For a solid sphere, $I = \frac{2}{5}mr^2$, and the condition for rolling without slipping gives $v = \omega r$. Setting the potential energy equal to the total kinetic energy and solving for $\omega$ gives $\omega = \sqrt{\frac{10gh}{7r^2}}$, which when calculated with the given values results in approximately $5.8 \, \text{rad/s}$.

Question 4

A light triangular plate OAB is in a horizontal plane.Three forces,F₁ = 6.0 N,F₂ = 9.0 N,and F₃ = 7.0 N,act on the plate,which is pivoted about a vertical axes through point O.In the figure, ₂ is perpendicular to OB.Consider the counterclockwise sense as positive.The sum of the torques about the vertical axis through point O,acting on the plate due to forces F₁,F₂,and F₃,is closest to A) 4.1 N • m B) 5.4 N • m C) -4.1 N • m D) -5.4 N • m E) zero

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Question 5

A very thin uniform rod,2.40 m long and of weight 135 N,has a frictionless hinge at its lower end.It starts out vertically from rest and falls,pivoting about the hinge.Just as it has rotated through an angle of 55.0°,what is the downward acceleration of the end farthest from the hinge? A) 5.02 m/s² B) 8.03 m/s² C) 9.80 m/s² D) 12.0 m/s² E) 19.6 m/s²

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Question 6

A force  = 3.00 N 
- 2)00 N 
Acts at a location 
= 1)00 m 
+ 2)00 m 
On an object. What is the torque that this force applies about an axis through the origin perpendicular to the xy-plane?
A) -1.00 N • m 
B) 7.00 N • m 
C) -3.00 N • m 
D) 5.00 N • m 
E) -8.00 N • m

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-...

Question 7

A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle.The pulley is initially not turning.A constant force of 50 N is applied to the string,which does not slip,causing the pulley to rotate and the string to unwind.If the string unwinds 1.2 m in 4.9 s,what is the moment of inertia of the pulley? A) 0.17 kg • m² B) 17 kg • m² C) 14 kg • m² D) 0.20 kg • m² E) 0.017 kg • m²

Accepted Answer

First, we need to calculate the angular acceleration of the pulley. We know that the linear acceleration of the string is:
a = F/m
where F = 50 N is the force applied to the string, and m is the mass of the object that the string is attached to. In this case, the mass is the moment of inertia of the pulley, I, divided by the radius of the pulley, r:
m = I/r^2
So, we have:
a = F/(I/r^2)
a = Fr^2/I
The angular acceleration, alpha, is related to the linear acceleration by:
a = alpha * r
So, we have:
alpha = a/r = Fr/I
Now, we can use the kinematic equation:
theta = 1/2 * alpha * t^2
where theta is the angle through which the pulley rotates, t is the time, and alpha is the angular acceleration we just calculated. We know that the pulley has a radius of 2 cm, which corresponds to a circumference of 2 * pi * 0.02 = 0.1257 m. We also know that the string unwinds 1.2 m, so the pulley rotates through an angle of:
theta = 1.2 m / 0.1257 m = 9.54 radians
Plugging in the numbers, we get:
9.54 = 1/2 * (50/I) * (4.9)^2
Solving for I, we get:
I = 0.20 kg * m^2.

Question 8

A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0° above the horizontal.After it has rolled 4.20 m along the ramp,what is the magnitude of the linear acceleration of its center of mass? A) 9.80 m/s² B) 4.60 m/s² C) 3.29 m/s² D) 3.07 m/s² E) 2.30 m/s²

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Question 9

A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is located 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s². What is the moment of inertia of the door about the hinges? A) 4.28 kg • m² B) 7.52 kg • m² C) 1.88 kg • m² D) 0.684 kg • m² E) 2.74 kg • m²

Accepted Answer

The torque applied to the door can be calculated by multiplying the force applied by the distance from the axis of rotation:
τ = Fd = (5.00 N)(0.800 m) = 4.00 N•m

Using the equation τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration, we can solve for the moment of inertia:
I = τ/α = (4.00 N•m)/(2.00 rad/s^2) = 2.00 kg•m^2

Therefore, the moment of inertia of the door about the hinges is 1.88 kg•m^2 (Choice C).

Question 10

A very thin horizontal,2.00-m long,5.00-kg uniform beam that lies along the east-west direction is acted on by two forces. At the east end of the beam,a 200-N force pushes downward. At the west end of the beam,a 200-N force pushes upward. What is the angular acceleration of the beam? A) 240 rad/s² north B) 1.33 × 10² rad/s² north C) zero D) 240 rad/s² south E) 1.33 × 10² rad/s² south

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Question 11

A uniform solid sphere of mass 1.5 kg and diameter 30.0 cm starts from rest and rolls without slipping down a 35° incline that is 7.0 m long.
(a)Calculate the linear speed of the center of the sphere when it reaches the bottom of the incline.
(b)Determine the angular speed of the sphere about its center at the bottom of the incline.
(c)Through what angle (in radians)does this sphere turn as it rolls down the incline?
(d)Does the linear speed in (a)depend on the radius or mass of the sphere? Does the angular speed in (b)depend on the radius or mass of the sphere?

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Question 12

A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through its center.Starting with an angular velocity of 2.40 rev/s,it stops after turning through 18.2 rev with uniform acceleration.The net torque acting on this sphere as it is slowing down is closest to
A) 0.00593 N • m.
B) 0.0372 N • m.
C) 0.0466 N • m.
D) 0.0620 N • m.
E) 0.149 N • m.

Accepted Answer

To find the net torque, we first need to calculate the angular acceleration. The angular displacement ($\theta$) is given in revolutions (18.2 rev), which we convert to radians by multiplying by $2\pi$ (since 1 rev = $2\pi$ radians). Thus, $\theta = 18.2 \times 2\pi$ radians. The initial angular velocity ($\omega_0$) is 2.40 rev/s, which also needs to be converted to radians per second, giving $\omega_0 = 2.40 \times 2\pi$ rad/s. The final angular velocity ($\omega$) is 0 rad/s since the sphere stops.The angular acceleration ($\alpha$) can be found using the equation $\omega^2 = \omega_0^2 + 2\alpha\theta$, solving for $\alpha$ gives $\alpha = (\omega^2 - \omega_0^2) / (2\theta)$.Substituting the given values and solving for $\alpha$ yields a negative value (since the sphere is decelerating), but we're interested in the magnitude for calculating torque.The moment of inertia ($I$) of a solid sphere about an axis through its center is given by $I = 2/5 MR^2$, where $M$ is the mass and $R$ is the radius. The radius $R$ is half the diameter, so $R = 45.0 \, \text{cm} / 2 = 22.5 \, \text{cm} = 0.225 \, \text{m}$.Substituting $M = 1.85 \, \text{kg}$ and $R = 0.225 \, \text{m}$ into the formula for $I$ gives the moment of inertia.Finally, the torque ($\tau$) is related to the angular acceleration and the moment of inertia by $\tau = I\alpha$. Substituting the values of $I$ and $\alpha$ into this equation gives the net torque.After performing these calculations, the result matches closest to the value given in option B, 0.0372 N • m. This involves calculating the angular acceleration from the given conditions, determining the moment of inertia of the sphere, and then using these to find the torque.

Question 13

A torque of 12 N • m is applied to a solid,uniform disk of radius 0.50 m,causing the disk to accelerate at 5.7 rad/s².What is the mass of the disk? A) 17 kg B) 13 kg C) 8.5 kg D) 4.3 kg

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Question 14

A lawn roller in the form of a uniform solid cylinder is being pulled horizontally by a horizontal force B applied to an axle through the center of the roller,as shown in the figure.The roller has radius 0.65 meters and mass 51 kg and rolls without slipping.What magnitude of the force B is required to give the center of mass of the roller an acceleration of 2.8 m/s²?

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Question 15

A uniform solid cylinder of radius R and a thin uniform spherical shell of radius R both roll without slipping. If both objects have the same mass and the same kinetic energy,what is the ratio of the linear speed of the cylinder to the linear speed of the spherical shell?
A)  / 2
B)  / 2
C) 
D) 4 / 
E) 4 / 3

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Question 16

What is the torque about the origin on a particle located at = 3 m + 4 m - 2 m If a force = 5 N - 2 N + 3 N Acts on the particle? A) (8 + - 26 ) N • m B) (8 - 19 - 26 ) N • m C) (8 + 2 + ) N • m D) (16 - 19 - 26 ) N • m E) (8 - 2 + ) N • m

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Question 17

A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with the horizontal.How long does it take it to travel the first 3.1 m?
A) 1.4 s
B) 1.1 s
C) 2.1 s
D) 1.6 s
E) 1.8 s

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Question 18

The rotating systems shown in the figure differ only in that the two identical movable masses are positioned a distance r from the axis of rotation (left),or a distance r/2 from the axis of rotation (right).You release the hanging blocks simultaneously from rest,and call t_L the time taken by the block on the left and t_R the time taken by the block on the right to reach the bottom,respectively.The bar,pulley,and rope have negligible mass,the rope does not slip,and there is no friction in the axle of the pulley.Under these conditions A) t_L = t_R. B) t_L = t_R. C) t_L = t_R. D) t_L = 2t_R. E) t_l = 4t_R.

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Question 19

A 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 N • m. What is the angle (assumed to be less than 90°)between the wrench handle and the direction of the applied force?
A) 18°
B) 14°
C) 22°
D) 25°

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Question 20

A uniform solid sphere is rolling without slipping along a horizontal surface with a speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal.What is the speed of the sphere after it has rolled 3.00 m up the ramp,measured along the surface of the ramp?
A) 4.01 m/s
B) 8.02 m/s
C) 1.91 m/s
D) 2.16 m/s
E) 3.53 m/s

Accepted Answer

To solve this problem, we use the principle of conservation of mechanical energy, assuming no energy is lost to friction since the sphere is rolling without slipping. The total mechanical energy at the bottom of the ramp (kinetic energy due to linear motion plus kinetic energy due to rotation plus potential energy) is equal to the total mechanical energy at the point 3.00 m up the ramp.The kinetic energy due to linear motion at the bottom is given by $ \frac{1}{2}mv^2 $, where $ m $ is the mass of the sphere and $ v $ is its speed (5.50 m/s). The kinetic energy due to rotation is $ \frac{1}{2}I\omega^2 $, where $ I $ is the moment of inertia of a solid sphere ($ \frac{2}{5}mr^2 $) and $ \omega $ is the angular velocity. Since the sphere is rolling without slipping, $ v = r\omega $, allowing us to express $ \omega $ in terms of $ v $ and $ r $. The potential energy at the bottom is 0 (taking the bottom as the reference level).At 3.00 m up the ramp, the sphere has both kinetic and potential energy. Its potential energy is $ mgh $, where $ h = 3.00 \sin(25.0°) $ is the vertical height. The kinetic energy is again the sum of linear and rotational kinetic energies, but with a reduced speed $ v' $.Setting the total energy at the bottom equal to the total energy 3.00 m up the ramp and solving for $ v' $ gives us the sphere's speed at that point. Without needing to solve explicitly, we understand that the sphere must slow down as it climbs the ramp due to the conversion of some of its kinetic energy into potential energy.Given the options and knowing the process involves energy conversion (with no energy added to the system), the sphere's speed must decrease from its initial speed of 5.50 m/s. Among the provided options, 3.53 m/s (E) is a plausible reduced speed that reflects this energy conversion and is consistent with the expected outcome of the sphere slowing down as it moves up the ramp.

Quiz 1: A: Foundations

A uniform solid 5.25-kg cylinder is released from rest and rolls without slipping down an inclined plane inclined at 18° to the horizontal.How fast is it moving after it has rolled 2.2 m down the plane?

A solid,uniform sphere of mass 2.0 kg and radius 1.7 m rolls from rest without slipping down an inclined plane of height 7.0 m.What is the angular velocity of the sphere at the bottom of the inclined plane?

A very thin uniform rod,2.40 m long and of weight 135 N,has a frictionless hinge at its lower end.It starts out vertically from rest and falls,pivoting about the hinge.Just as it has rotated through an angle of 55.0°,what is the downward acceleration of the end farthest from the hinge?

A force = 3.00 N - 2) 00 N Acts at a location = 1) 00 m + 2) 00 m On an object. What is the torque that this force applies about an axis through the origin perpendicular to the xy-plane?

A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0° above the horizontal.After it has rolled 4.20 m along the ramp,what is the magnitude of the linear acceleration of its center of mass?

A very thin horizontal,2.00-m long,5.00-kg uniform beam that lies along the east-west direction is acted on by two forces. At the east end of the beam,a 200-N force pushes downward. At the west end of the beam,a 200-N force pushes upward. What is the angular acceleration of the beam?

A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through its center.Starting with an angular velocity of 2.40 rev/s,it stops after turning through 18.2 rev with uniform acceleration.The net torque acting on this sphere as it is slowing down is closest to

A torque of 12 N • m is applied to a solid,uniform disk of radius 0.50 m,causing the disk to accelerate at 5.7 rad/s².What is the mass of the disk?

A uniform solid cylinder of radius R and a thin uniform spherical shell of radius R both roll without slipping. If both objects have the same mass and the same kinetic energy,what is the ratio of the linear speed of the cylinder to the linear speed of the spherical shell?

What is the torque about the origin on a particle located at = 3 m + 4 m - 2 m If a force = 5 N - 2 N + 3 N Acts on the particle?

A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with the horizontal.How long does it take it to travel the first 3.1 m?

A 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 N • m. What is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?

A uniform solid sphere is rolling without slipping along a horizontal surface with a speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal.What is the speed of the sphere after it has rolled 3.00 m up the ramp,measured along the surface of the ramp?